The area (in sq. units) bounded by the parabola y2 = 8x and the line x = 2 is

$\frac{16}{3}$ $\frac{32}{3}$ $\frac{32 \sqrt{2}}{3}$ $\frac{16 \sqrt{2}}{3}$

$\frac{32}{3}$

y2 = 8x so $y = 2\sqrt{2}\sqrt{x}$ x > 2       area in 1st quadrant = area of 4th quadrant ⇒  area in 1st quadrant × 2 $=2 \times \int\limits_0^2 2 \sqrt{2} \sqrt{x} d x$ $=4 \sqrt{2}\left[\frac{2 x^{3 / 2}}{3}\right]_0^2=\frac{4 \sqrt{2} \times 2 \times 2 \sqrt{2}}{3}$ ⇒  $\frac{32}{3}$ sq. units